Is there a limit for voltage a PM generator gets to based on RPMs? - Electrical Engineering Stack Exchange most recent 30 from electronics.stackexchange.com 2019-07-24T01:08:53Z https://electronics.stackexchange.com/feeds/question/409922 http://www.creativecommons.org/licenses/by-sa/3.0/rdf https://electronics.stackexchange.com/q/409922 0 Is there a limit for voltage a PM generator gets to based on RPMs? Robert Dold https://electronics.stackexchange.com/users/157932 2018-12-01T17:10:52Z 2018-12-02T10:18:06Z <p>I build different small PM handcrank generators for fun, Recently I built a single phase generator and printed myself a couple transmissions. Now as far as I know 3 things affect voltage, speed of rotation, coil size/#of windings, and magnetic strength. Where I am mentally stuck is that being the same generator (all things equal) when I ran it on my 1-250 transmission, i generated 9 volts with a given load, with the same load and generator, using the 1-400, I got the SAME output, even though I was going faster (i have an rpm reader). Up until now, all the generators I have bought, many of them wye 3 phase, the faster i go, the more voltage. I am certain I am not reaching the limit of my wire, as im using 22 gauge, and only pushing 2.5 amps max. While there is a possibility this was a testing error, I double checked and got the same results. Has anyone ever run into this Voltage wall before? as in no matter how fast you rotate, you hit a wall. </p> https://electronics.stackexchange.com/questions/409922/-/410017#410017 2 Answer by Transistor for Is there a limit for voltage a PM generator gets to based on RPMs? Transistor https://electronics.stackexchange.com/users/73158 2018-12-02T10:18:06Z 2018-12-02T10:18:06Z <p>As discussed in the comments, you need to consider the winding inductance. As you increase the frequency the impedance will increase as given by the equation <span class="math-container">\$Z=jωL = 2 \pi fL \$</span>. That means that there will be an increase in the internal impedance of your coils and they will act as a potential divider with the load.</p> <p>You can do make a reasonable measurement with your multimeter as follows:</p> <ul> <li>Take an open circuit true RMS voltage measurement from your generator.</li> <li>Connect up to your load and find what frequency causes the output to be half the open-circuit voltage.</li> <li>Repeat step one at that frequency and check step two again to confirm that the AC reading is halved when on load.</li> </ul> <p>Now you have the frequency where the impedance of the coil matches that of the load, R. From that we can say <span class="math-container">$$R = 2 \pi f L$$</span> so <span class="math-container">$$L = \frac {R}{2 \pi f}$$</span></p> <p>For a more thourough explanation see <a href="https://electronics.stackexchange.com/questions/220672/is-it-possible-to-measure-inductance-without-lcr-meter-osc-fg">Is it possible to measure inductance without LCR Meter/OSC/FG</a>.</p> <p>With the coil's inductance and resistance measurements you can do all sorts of calculations and predict the output to a given load at various speeds.</p>